Book Chapter Lag Rang Ian

7/31/2019 Book Chapter Lag Rang Ian

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Chapter 22

The Hamiltonian and Lagrangian densities

from my book:

Understanding Relativistic Quantum Field Theory

Hans de Vries

January 2, 2009


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2 Chapter


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Contents

22 The Hamiltonian and Lagrangian densities 1

22.1 The relativistic Hamiltonian and Lagrangian . . . . . . . . 222.2 Principle of least action / least proper time . . . . . . . . 422.3 The Hamiltonian and Lagrangian density . . . . . . . . . 622.4 The Euler-Lagrange equation for Fields . . . . . . . . . . . 722.5 Lagrangian of the scalar Klein Gordon field . . . . . . . . 922.6 Hamiltonian of the scalar Klein Gordon field . . . . . . . . 1122.7 The complex Klein Gordon field . . . . . . . . . . . . . . . 1222.8 Expressing the Lagrangian in a 4d environment . . . . . . 14

22.9 Electromagnetic and Proca Langrangian . . . . . . . . . . 1522.10 The electromagnetic equation of motion . . . . . . . . . . 1822.11 The electromagnetic interaction Lagrangian . . . . . . . . 1922.12 Electromagnetic and Proca Hamiltonian . . . . . . . . . . 20


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Chapter 22

The Hamiltonian andLagrangian densities


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2 Chapter 22. The Hamiltonian and Lagrangian densities

22.1 The relativistic Hamiltonian and Lagrangian

The Hamiltonian and Lagrangian which are rather abstract constructionsin classical mechanics get a very simple interpretation in relativistic quan-tum mechanics. Both are proportional to the number of phase changes perunit of time. The Hamiltonian runs over the time axis while the Lagrangianruns over the trajectory of the moving particle, the t-axis.

Figure 22.1: The Hamiltonian and Lagrangian

Figure 22.1 shows the relativistic de Broglie wave in a Minkowski dia-gram. The triangle represents the relation between the Lagrangian an theHamiltonian, which holds in both relativistic and non-relativistic physics.

L = pv H (22.1)

The Hamiltonian counts the phase-changes per unit of time on the verticalaxis while the term pv counts the phase-changes per unit on the horizontalaxis: v is the distance traveled per unit of time while p is proportionalwith the phase-changes per unit of distance, hence the term pv. We cannow understand the classical relation. (with q = x = v)

L

q= p (22.2)


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22.1 The relativistic Hamiltonian and Lagrangian 3

For the free classical relativistic particle we have for the Hamiltonian (En-

ergy) and the pv term.

H =mc21 v

2

c2

, pv =mv21 v

2

c2

(22.3)

Calculating the Lagrangian we see that the Hamiltonian is proportional to while the Lagrangian is proportional to 1/.

L = (H pv) =

c2 v21 v

2

c2m =

1

v2

c2 mc2

(22.4)

This is what we expect from time dilation. The moving particle has lessclock-ticks by a factor due to the time dilation, We now check that.

L

q=

v

1

v2

c2mc2

=

mv1 v

2

c2

= p (22.5)

For sofar we have not yet discussed the potential energy. To obtain theequation of motion of a relativistic particle in a potential field we have toadd the potential energy term V(q). In the non-relativistic case we have.

H = T(q) + V(q), L = T(q) V(q) (22.6)

Where T(q) = 12

mv2 is the kinetic energy. The relativistic Hamiltonianand Lagrangian we have discussed however also include the restmass en-

ergy. The restmass energy can be considered as being part of the potentialenergy. The kinetic part T in the relativistic case can be obtained asfollows.

H + L = 2T = pv

T =1

2pv =

1

2

mv21 v

2

c2

1

2mv2 ( for v c ) (22.7)

Using the term L = 12

mv2 in eq. (22.5) gives us p = mv for the non-

relativistic momentum


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22.2 Principle of least action / least proper time

To obtain the relativistic equation of motion of a particle in a potential fieldwe use Lagrange equation of motion derived from Hamiltons variationalprinciple of Least Action. Relativistic quantum mechanics gives us thedeeper perspective of this principle which now becomes the principle ofleast phase-change and thus the least proper time: The relativistic particlefollows the path which will bring it there in the shortest proper time.

We will briefly recall the derivation here, it is one of the most fundamental

principles of physics. Somewhat abstract we can write.

S =

t2t1

L(q, q) dt = 0 (22.8)

Where S is the variation of the action, the variation of the proper timein quantum mechanics, is zero, meaning that we have a minimum or max-imum just like the a zero first order derivative of a function indicates alocal minimum or maximum. The particle traverses a path between twofixed points x1 and x2 between t1 and t2 which are also fixed. We assumethat the Lagrangian only depends on the position q of the particle and itsvelocity q. Now we allow small arbitrary variations in q and q along thepath.

S =

t2t1

L

qq +

L

qq

dt = 0 (22.9)

As long as these variations are small we can determine the change in Lwith the help of the first order derivatives. The last term above is alsothe last term in the expression below according to the product rule fordifferentiation.

d

dt

L

qq

=

d

dt

L

q

q +

L

q

d

dt

q

(22.10)

We can instantly integrate the left hand term which gives us the following.

S =t2t1

Lq

ddt

Lq

q dt +

Lq q t2

t1

= 0 (22.11)


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22.2 Principle of least action / least proper time 5

The integrated term at the end vanishes since q is defined as zero at theend-points. Since q is arbitrary along the rest of the path we have to setthe term between brackets to zero. This then gives us the Euler-Lagrangeequation.

d

dt

L

q

L

q= 0 (22.12)

It defines a local minimum at each point along the path. We can usethe metaphor of a ball rolling at the bottom of a valley. Inserting theLagrangian of the classical relativistic particle.

L =

1

v2

c2mc2 V(q) (22.13)

Gives us the equation of motion of the relativistic particle in a potentialfield.

dp

dt=

dV

dq, with p =

mv1 v2/c2

(22.14)

A gradient of the potential field causes a change in the relativistic momen-tum: If we use the non-relativistic Lagrangian with,

L 1

2mv2 V(q) (22.15)

then we get the non-relativistic equation of motion.

ma = dV

dq(22.16)

As long as the derivation is correct for the relativistic particle, then we canbe assured that it is valid for the non-relativistic limit as well.


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22.3 The Hamiltonian and Lagrangian density

We can define the Hamiltonian and Lagrangian density for any extendedobject, being either classical or a quantum field, as.

H =

H dx3, L =

L dx3 (22.17)

Let us see how these quantities transform under Lorentz transformation.We did see in (22.3) and (22.4) that the integrated quantities, the Hamil-

tonian H, the Lagrangian L and the term pv transform like.

H transforms as pv transforms as 2L transforms as 1/

(22.18)

The volume of a wave-functions transforms like 1/ due to Lorentz con-traction. So, the densities become higher by a factor , hence the densityH, the Lagrangian density L and the density of the pv term transform like.

H transforms as 2

PV transforms as 22

L transforms as 1(22.19)

We see that the Lagrangian density is the same in all reference frames.It is a Lorentz scalar. This makes the Lagrangian density a fundamentalquantity in quantum field theory. The Standard Model of physics is basedon the Lagrangian density which in quantum physics is generally called

just the Lagrangian, without the density.

The equations of motion are based on the derivatives of the Lagrangiandensity which is a Lorentz scalar. Done in the right way assures that thewhole Standard Model of physics transforms in the right way, that is, thelaws of physics are the same in every reference frame.

The triangular equation H pv = L basically counts phase changeclock-pulses on the t, x and t-axis. The corresponding relation of thedensities transforms like the basic energy/momentum relation.

H PV = L transforms as E2 p2 = m2 (22.20)


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22.4 The Euler-Lagrange equation for Fields 7

22.4 The Euler-Lagrange equation for Fields

The Euler-Lagrange equation for fields operates on a lagrangian whichdepends only on the (generalized) coordinate q and velocity q of the par-ticle. It is valid for relativistic particles even though it was developed byLeonhard Euler and Joseph-Louis Lagrange in the 1750s.

L L(q, q) (22.21)

In quantum field theory we do not have a quantity like q explicitly available.We work with a field (t, r) instead. We assume that the Lagrangian

density only depends on (t, r) and its first order derivatives.

L L(, t, x, y, z) (22.22)

We will avoid in this book the widespread custom to present the correctrelativistic Lagrangian density for the (scalar) Klein Gordon field and then

justify it by making substitutions like.

q ,1

2 mv2

1

2 2

(22.23)

The term 12

mv2 should worry the reader. Indeed the Hamiltonian densitysubsequently derived does transform in the wrong way and its integral overspace does not correspond with the Hamiltonian of the classical relativisticparticle.

The origin of these substitutions can be understood by looking at ourinitial mechanical spring-mass model of the Klein Gordon equation shownin figure ??. In this model is a displacement which could be associatedwith q. These substitutions however are to naive, worse, they lead toviolations of special relativity.

Rather then trying to make associations between classical and field termswe want to stress the fact that the Euler-Lagrange mechanism to derive theequations of motion is an entirely mathematical mechanism to find a min-imum/maximum. It does not matter by what quantities the Lagrangiandensity is expressed as long as they express the right one.

The Euler-Lagrange equation for quantum fields goes well beyond thescalar Klein Gordon field. It holds for all quantum fields fields. The


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derivation is the same as the derivation for the classical one. We will

follow the same four steps as in equations (22.9) through (22.13) for theclassical wave equation. The variation of the action is symbolized by.

S =

t2t1

dt

+

L(, t, x, y, z) dx3 = 0 (22.24)

More concretely we can write it expressed in variations of the field andits (four) derivatives.

S =

t2

t1

L

+

L

())

dx4 = 0 (22.25)

The last term above is equal to the last term in the equation below whichjust expresses the product rule for differentiation. In fact there are fourterms in total, one for each derivative.

L

() =

L

() +L

()

(22.26)

The left hand term above (representing four terms) can be directly inte-grated over one axis, each of the four along its own axis. The result ofthese integrations is zero since the variations at the end points are definedas zero. So, we can omit the integration over the other three axis and con-tinue with the remaining terms which are all proportional to the variationof itself.

S =

t2

t1

L

L

()

dx4 = 0 (22.27)

Since is totally arbitrary we conclude that the equation between brack-ets has to hold at each point (This means we have nothing to do anymorewith where the borders are in which reference frame). We have obtainedthe Euler Lagrange equation for the relativistic Lagrangian density:

L

() L

= 0 (22.28)


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22.5 Lagrangian of the scalar Klein Gordon field 9

22.5 Lagrangian of the scalar Klein Gordon field

The form of the Hamiltonian and Lagrangian densities of the Klein Gordonfield are determined by the fact the Klein Gordon field is a scalar field. Thismeans that the values of the field are Lorentz invariant. They are thesame in any reference frame.

If we perform a Lorentz transform on the field like (x) = (x) thenis suffices to transform the coordinates x to obtain (x)

(x

) = (x

) = (

1

x

) (22.29)

We did see that the Hamiltonian density transforms as E2, while the kinetic(density) term T transforms like p2, see (22.20). Since the scalar field itself doesnt transform we might expect differential operators which obtainE2 and p2 from the quantum field instead. We will see that this is indeedthe case.

In the classical non relativistic theory the Lagrangian is given by L = TV.

In case of a relativistic free particle there is no potential energy, howeverthere is the energy corresponding with the mass of the particle. This self-energy term, which plays a somewhat similar role as the potential energy,is absent in the non-relativistic theory.

We did see that the kinetic term T becomes 12pv in the relativistic case.

We want to write the relativistic case in a form of L = T W, where Wrelates to the mass-energy. Using the generally valid L = pv H we canrewrite L like:

L = 12

H + 12

pv + 12

L (22.30)

For a classical relativistic particle we should substitute the right handterms as follows (see section 22.1)

L = 1

2mc2 +

1

2mv2

1

21mc2 (22.31)

Going from the Lagrangian to the Lagrangian density we have to multiplythe right hand terms with and extra factor to compensate for Lorentz


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contraction which confines the wave-function into a smaller volume, hence

the density goes up by a factor . At this point we will also multiply allright hand terms with the constant mc2 for convenance. We will see why.So, all the right hand terms are multiplied by mc2 when going from thelagrangian L to the Lagrangian density L.

L = 1

2

mc2

2+

1

2[mv ]2 c2

1

2

mc2

2(22.32)

The terms between square brackets we recognize as E, p and the rest-mass

energy. We can now make the step from the classical relativistic particleto the Klein Gordon field theory:

L =1

2

t

2

1

2

xi

2c2

1

2

mc2

2(22.33)

Setting (c = = 1) gives us the familiar form of the Lagrangian density ofthe Klein Gordon field.

L =

1

2

2

1

2

1

2 m

2

2

(22.34)

Note that the middle term at the right hand side corresponds with theclassical kinetic term T which becomes the kinetic energy in the non rel-ativistic theory of the classical particle. We have derived the Lagrangiandensity for the scalar field. By assuming that is a Lorentz scalar weobtained the above Lagrangian density. We now use the Euler Lagrangeequation.

L

()

L

= 0 (22.35)

In order to obtain the equation of motion. What we get is the Klein Gordonequation.

2

t2

2

x2i+ m2 = 0 (22.36)

The scalar quantum field representation derived directly from the classicalrelativistic particle Lagrangian.


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22.6 Hamiltonian of the scalar Klein Gordon field 11

22.6 Hamiltonian of the scalar Klein Gordon field

For the Hamiltonian density we go back to the expression for the classicalrelativistic particle since we want both Hamiltonians to correspond. Usingequation (22.32) and H = 2T L we get for the Hamiltonian density ofthe classical particle.

H =1

2

mc2

2+

1

2[mv ]2 c2 +

1

2

mc2

2(22.37)

This corresponds with the following Hamiltonian density for the KleinGordon equation.

H = 1

22

1

2 +

1

2m22 (22.38)

With (c = = 1). We can see how this Hamiltonian density transformsby applying it to a plane-wave of the form exp(iEt + ipx). We get:

H E2 + p2 + m2 2 + 22 + 1 2 (22.39)

Thus: the Hamiltonian density transforms like 2 where one factor stemsfrom the Hamiltonian being the 0th component of the 4-momentum andthe second factor comes from the Lorentz contraction of the volumewhich confines the field.

This Hamiltonian density corresponds with the classical particle but differsin signs with the second quantization related Hamiltonian density1

1The Hamiltonian related with the second quantization of the Klein Gordon fieldis given by H = 1

22 + 1

2( )2 + 1

2m22. The reason of the difference in signs

is that the term is considered as the momentum in an internal or unspecifiedspace. The term 1

22 is then considered to be the T in H = 2T L. The general

problem in applying second order quantization in the relativistic theory is that the mixof internal or unspecified space and the usual space-time coordinates doesnt lead

to the correct Lorentz transformation. A second problem is that 12 2 corresponds to thenon-relativistic expression 1

2mv2 instead of the relativistic version 1

2pv


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22.7 The complex Klein Gordon field

The complex Klein Gordon equation comes in when we need to describeboth particles and anti-particles. becomes a field with two components1 and 2.

= 1 + i2 (22.40)

How can we interpret these components. One possibility is to take one

component as a position (in some internal or unspecified space andthe other component as the momentum. The two would be 90o out ofphase in an oscillatory motion which we could associate with the particlesfrequency exp(iEt/)

Another way is to interpret both as coordinates on a plane of rotation. Theexpression exp(iEt/) would then correspond with a circular motion2.

The Lagrangian for the complex field must contain both components andthe Euler-Lagrange equation must lead to the usual Klein Gordon equation.

Say we us the Lagrangian of the real Klein Gordon equation,

L =1

22

1

2

1

2m22 (22.41)

and simply use as a complex variable. Evaluating this with a plane wavesolution however doesnt produce a real value. L contains the local phaseof the wave-function, its not a real value.

L = 12

E2 +1

2p2

1

2m2 2 = m22 (22.42)

The result we want is m2||2 or m2 which more explicitly expressedin its individual components is m2

2 + 2

without the imaginary val-

ued cross-term 2im212 included in m22. We get the Lagrangian

density we want by simply adding the two Lagrangians of the individualcomponents together. The same can be done for the Hamiltonian.

2Both interpretations lead to a specific direction in space, either the direction ofoscillation or the spin pointer. This is an issue for a scalar theory such as the one

represented by the scalar Klein Gordon equation, which is not supposed to have anyspecial direction in space


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22.7 The complex Klein Gordon field 13

Lagrangian density for the complex scalar field 1 + i2

L =

+

1

221

1

21 1

1

2m221

+

1

222

1

22 2

1

2m222

(22.43)

L =1

2

+ m2

(22.44)

Hamiltonian density for the complex scalar field 1 + i2

H =

+

1

221

1

21 1 +

1

2m221

+

1

222

1

22 2 +

1

2m222

(22.45)

H(x) =1

2

+ + m2

(22.46)

Some care is required with the signs here since = (21+22). Equation

(22.44) is often found with reversed signs. One can easily check the requiredsigns by inserting a plane-wave eigenfunction into the Lagrangian density: = 1 + i2 = cos(Et + px) + i sin(Et + px), One should get.

L =1

2

E2 + p2 m2

= m2 (22.47)

The equation of motion can be derived in a mathematically proper way3 by

applying the Euler-Lagrange equation on (22.43), taking the derivatives inthe fields 1, 2 and their derivatives, and then defining the combined fieldagain as = 1 + i2. The result is the familiar Klein Gordon equation.

2

t2

2

x2i= m2 (22.48)

3Despite what is often seen, the Euler Lagrange equation can not be applied directlyon expressions containing terms like . A derivative like / is not zero but un-

determined since it violates the Cauchy-Riemann equations for complex differentiability:The derivative is not independent of the direction of in the complex plane.


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22.8 Expressing the Lagrangian in a 4d environment

The reader may have noticed that there seems to be an ambiguity in how wedefine the Lagrangian density. The expression for the Lagrangian densitywe obtained is.

L =1

2

2 +

2 m22

(22.49)

While we in fact could also have written.

L = m22 or L = 2 +

2 (22.50)

Both expressions lead to the amount of phase change per unit of time perunit of volume over the trajectory of the particle, at least for plane-waves.Instead we end up with a linear combination of the latter two expressions.Why? For the Hamiltonian it seems that we could equally well write.

H = 2 or H = 2

+ m22 (22.51)

To get the amount of phase change per unit of time per unit of volumeover the time-axis. Instead we ended up with.

H(x) =1

2

2 +

2+ m22

(22.52)

One important argument is that we do not only want to know the La-grangian or Hamiltonian density in one particular reference frame, but we

want to know these quantities in all reference frames. Knowing a phasechange rate in one particular space-time direction doesnt say anythingabout the other directions. We need sufficient information to be ableto transform the Lagrangian and Hamiltonian density into any referenceframe.

One might suspect that Nature itself also needs such a definition in 4dspace-time, and that therefor, even though the expressions seem to beambiguous from a single reference point of view, they actually do representthe physics as required in a 4d-dimensional world obeying the rules of

special relativity.


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22.9 Electromagnetic and Proca Langrangian 15

22.9 Electromagnetic and Proca Langrangian

In correspondence with the Lagrangian densities discussed sofar we mightexpect the Lagrangian density for the electromagnetic four-vector to beexpressed by the following.

Lp =

+

1

2A20

1

2A0 A0

1

2m2A20

12A2x 12Ax Ax 12 m2A2x

1

2A2y

1

2Ay Ay

1

2m2A2y

1

2A2z

1

2Az Az

1

2m2A2z

(22.53)

Where all four components ofA have independently the form of the (clas-sical) Lagrangian field density. The requirement that the total Lagrangiandensity transforms like a Lorentz scalar imposes the (+,-,-,-) metric on the

time/space components, the signs in the first column of (22.53). Theseexpressions can be written more compact as:

Lp =1

2A A

1

2m2c2AA (22.54)

If the mass m is not zero then we call the field a Proca field. This expressionis however not complete. We have to replace the derivatives of A in thefollowing sense. (Well discuss the reason for this in a minute)

A = A A, A = A A (22.55)

So the Lagrangian density becomes. (in the massless case)

L =1

2(A A) (A A) = F

F (22.56)

Which we can write (using the normalization in SI) as.


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L = 14

FF = 12

BH - DE

(22.57)

This Lagrangian density is zero in case of electromagnetic radiation wherethe relation |E| = |cB| holds always. This shows us that the invariantphoton mass is zero. (The partial Lagrangian density (22.54) is zero aswell in this case)

Now, why do we need to subtract these extra terms? Well typically the

energy-momentum of the electromagnetic field is derived by calculatinghow the field acts on charges on a capacitor and currents in an inductor,so charge is involved in one way or the other and charge is represented byan U(1) symmetry: exp(i).

The terms we need to subtract induce an equal and indistinguishable U(1)phase as the regular Lagrangian components and therefor need to be takencare of. The total induced phase on a charged scalar field by the four-vector A is defined in the following way.

= exp

i

(po + eAo)dxo +

i

3i=1

(pi + eAi)dxi

(22.58)

Where p is the inertial momentum defined by the invariant mass. We cansplit of the factor dependent on A as follows = p A. Since is ascalar we can write = , the order of the differential operatorsdoes not matter.

The combination (p + eA) must be curl free in any of the six 2D planesof 4D space-time otherwise otherwise the expression within the squarebrackets can not be a single valued scalar function.

(p + eA) ds = 0 (22.59)

The individual terms p and eA can have curl, so any curl in eA must

be canceled by an opposite curl in p. This relationship gives rise tothe electromagnetic Lorentz Force. The expression i p yields the


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22.9 Electromagnetic and Proca Langrangian 17

changes of the momenta p in the directions x. All of these terms end up

in the expression which describes the change of momentum of a particlemoving at some speed v since.

dp

dt=

p

t+

p

xvx +

p

yvy +

p

zvz (22.60)

Where the right hand side is just the mathematical expansion of the lefthand side (vx = x/t, et-cetera ) It is no surprise that the expressionA gives rise to the terms which end up in the electromagnetic field

tensor F which yields the changing momentum.

dp

dt= e

A A

v = e Fv (22.61)

Where the term between brackets represents the curl of A. Since theorder of differentiation is irrelevant we can deduce.

=

=

(p

+ eA

) =

(p

+ eA

) (22.62)

The righthand side can be reordered as.

p p = e ( A A ) (22.63)

This expression simplifies for a particle with v = 0, and thus p = 0 (at

every point in space) to.

pi

xo= e

Ai

xo e

Ao

xi= e E (22.64)

Which is just the electric part of the Lorentz force. A non-zero velocityv, constant over space, gives rise to the full Lorentz force including themagnetic terms.

We see that the assumption that somehow charge is involved in the La-grangian leads to extra terms which are indistinguishable from the basic


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terms because of the U(1) symmetry. Certainly a term like 12

DE suggests

likewise. It is the energy needed for the vacuum displacement current Dto move through the electric field E which builds up in the process as aresult.

22.10 The electromagnetic equation of motion

We can derive the equations of motion with the use of the Euler Lagrangeequation starting from the Lagrangian density.

L =1

4o

AA

AA

=

1

2o

B2

1

c2E2

(22.65)

Multiplying the terms and realizing that a nd are dummy variables toadd all terms together to a single scalar result, we can rewrite this in theform which is generally used to derive the equations of motion.

L =1

2o A A A A (22.66)

The first of the two terms is identical to the initial Lagrangian density(22.53) while the second term corresponds to the extra terms. We recallthe Euler Lagrange equation.

L

(A)

L

A= 0 (22.67)

Applying it gives us the equation of motion.

A A

= 0 (22.68)

The term A is zero if A is a conserved current. We see that the

equation of motion is not changed by the extra terms if this is the case. Incase of a vacuum without net charge-current density we obtain.

A = 0 (22.69)


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22.11 The electromagnetic interaction Lagrangian 19

22.11 The electromagnetic interaction Lagrangian

The electromagnetic interaction Lagrangian density represents the extraphase change rates induced by the four-potential A on a charged (KleinGordon) field . It is given by.

Lint = jA (22.70)

Which transforms as it should like a Lorentz scalar. This gives us for thetotal equation of motion.

A A

= oj (22.71)

Which becomes under the assumption that A is a conserved current(Lorentz gauge condition).

A = oj

(22.72)

Here we recover the classical the classical wave equation of A with thecharge-current density j as its source. We can write the equation ofmotion as.

A A

= oj

(22.73)

This allows us to express it with the use of F instead of A.

F = oj (22.74)

Which is a compact way of writing of the inhomogeneous Maxwell equa-tions, which we have hereby derived from the Lagrangian density.

E = cojo =

1

o (22.75)

B 1

c2

E

t

= oj (22.76)


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22.12 Electromagnetic and Proca Hamiltonian

Well follow the same steps here as we did for the electromagnetic La-grangian density. In correspondence with the Hamiltonian densities dis-cussed sofar we expect the Hamiltonian density for the electromagneticfour-vector to be expressed by the following.

Hp =

+

12

A20 1

2A0 A0 +

1

2m2A20

+

1

2A2x

1

2Ax Ax +1

2 m2

A2x

+

12

A2y 1

2Ay Ay +

1

2m2A2y

+

12

A2z 1

2Az Az +

1

2m2A2z

(22.77)

Where all four components of A have independently the form of the (clas-sical) Hamiltonian field density. These expressions can be written more

compact as:

Hp = 1

2A A +

1

2mc2AA (22.78)

We used instead of here for later convenience. Both give the sameend result due to the square. This expression is again not complete. Wehave to replace the derivatives of A in the following sense.

A = A A, (22.79)

For the same reason as explained in the section on the Lagrangian density.With a zero mass field normalized for SI units this leads us to the wellknow expression for the energy density of the electromagnetic field.

H =1

4o

A A

A A

(22.80)

H =1

4oFF =

1

2BH + DE (22.81)

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